details

property	value
status	complete
benchmark	test76.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n089.star.cs.uiowa.edu
space	Strategy_removed_mixed_05

run statistics

property	value
solver	NTI_22
configuration	default
runtime (wallclock)	0.664036989212 seconds
cpu usage	1.750276895
max memory	4.2534912E8

stage attributes

key	value
output-size	3805
starexec-result	NO

output

/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- NO ** BEGIN proof argument ** The following rule was generated while unfolding the analyzed TRS: [iteration = 5] f(g(0,_0),s(0),g(0,_0)) -> f(g(0,_0),_0,g(0,_0)) Let l be the left-hand side and r be the right-hand side of this rule. Let p = epsilon, theta1 = {_0->s(0)} and theta2 = {}. We have r|p = f(g(0,_0),_0,g(0,_0)) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = f(g(0,s(0)),s(0),g(0,s(0))) loops w.r.t. the analyzed TRS. ** END proof argument ** ** BEGIN proof description ** ## Searching for a generalized rewrite rule (a rule whose right-hand side contains a variable that does not occur in the left-hand side)... No generalized rewrite rule found! ## Applying the DP framework... ## 2 initial DP problems to solve. ## First, we try to decompose these problems into smaller problems. ## Round 1 [2 DP problems]: ## DP problem: Dependency pairs = [f^#(0,s(0),_0) -> f^#(_0,+(_0,_0),_0)] TRS = {+(_0,0) -> _0, +(_0,s(_1)) -> s(+(_0,_1)), f(0,s(0),_0) -> f(_0,+(_0,_0),_0), g(_0,_1) -> _0, g(_0,_1) -> _1} ## Trying with homeomorphic embeddings... Failed! ## Trying with polynomial interpretations... This DP problem is too complex! Aborting! ## Trying with lexicographic path orders... Failed! ## Trying with Knuth-Bendix orders... Failed! Don't know whether this DP problem is finite. ## DP problem: Dependency pairs = [+^#(_0,s(_1)) -> +^#(_0,_1)] TRS = {+(_0,0) -> _0, +(_0,s(_1)) -> s(+(_0,_1)), f(0,s(0),_0) -> f(_0,+(_0,_0),_0), g(_0,_1) -> _0, g(_0,_1) -> _1} ## Trying with homeomorphic embeddings... Success! This DP problem is finite. ## A DP problem could not be proved finite. ## Now, we try to prove that this problem is infinite. ## Trying to find a loop (forward=true, backward=true, max=20) # max_depth=20, unfold_variables=false: # Iteration 0: no loop found, 1 unfolded rule generated. # Iteration 1: no loop found, 1 unfolded rule generated. # Iteration 2: no loop found, 7 unfolded rules generated. # Iteration 3: no loop found, 65 unfolded rules generated. # Iteration 4: no loop found, 805 unfolded rules generated. # Iteration 5: success, found a loop, 6983 unfolded rules generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = f^#(0,s(0),_0) -> f^#(_0,+(_0,_0),_0) [trans] is in U_IR^0. We build a unit triple from L0. ==> L1 = f^#(0,s(0),_0) -> f^#(_0,+(_0,_0),_0) [unit] is in U_IR^1. Let p1 = [0]. We unfold the rule of L1 backwards at position p1 with the rule g(_0,_1) -> _0. ==> L2 = f^#(g(0,_0),s(0),g(0,_0)) -> f^#(g(0,_0),+(g(0,_0),g(0,_0)),g(0,_0)) [unit] is in U_IR^2. Let p2 = [1, 0]. We unfold the rule of L2 forwards at position p2 with the rule g(_0,_1) -> _1. ==> L3 = f^#(g(0,_0),s(0),g(0,_0)) -> f^#(g(0,_0),+(_0,g(0,_0)),g(0,_0)) [unit] is in U_IR^3. Let p3 = [1, 1]. We unfold the rule of L3 forwards at position p3 with the rule g(_0,_1) -> _0. ==> L4 = f^#(g(0,_0),s(0),g(0,_0)) -> f^#(g(0,_0),+(_0,0),g(0,_0)) [unit] is in U_IR^4. Let p4 = [1]. We unfold the rule of L4 forwards at position p4 with the rule +(_0,0) -> _0. ==> L5 = f^#(g(0,_0),s(0),g(0,_0)) -> f^#(g(0,_0),_0,g(0,_0)) [unit] is in U_IR^5. This DP problem is infinite. Proof run on Linux version 3.10.0-1160.25.1.el7.x86_64 for amd64 using Java version 1.8.0_292 ** END proof description ** Total number of generated unfolded rules = 7913 popout

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job log

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actions

all output return to TRS Standard